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A321196 Riordan triangle T = (1/(1 + x^2 - x^3), x/(1 + x^2 - x^3)).

%I A321196 #25 Mar 24 2020 03:23:36
%S A321196 1,0,1,-1,0,1,1,-2,0,1,1,2,-3,0,1,-2,3,3,-4,0,1,0,-6,6,4,-5,0,1,3,-1,
%T A321196 -12,10,5,-6,0,1,-2,12,-4,-20,15,6,-7,0,1,-3,-7,30,-10,-30,21,7,-8,0,
%U A321196 1,5,-16,-15,60,-20,-42,28,8,-9,0,1
%N A321196 Riordan triangle T = (1/(1 + x^2 - x^3), x/(1 + x^2 - x^3)).
%C A321196 This is the (ordinary) convolution triangle based on A077961 (the column k = 0 of T).
%C A321196 The row polynomials R(n, x) := Sum_{k=0..n} T(n, k)*x^k, with R(-1, x) = 0, appear in the Cayley-Hamilton formula for nonnegative powers of a 3 X 3 matrix with Det M = sigma(3; 3) = x1*x2*x3 = +1, sigma(3; 2) := x1*x2 + x1*x*3 + x2*x^3 = +1 and Tr M = sigma(3; 1) = x1 + x2 = x, where x1, x2, and x3 are the eigenvalues of M, and sigma the elementary symmetric functions, as M^n = R(n-2, x)*M^2 + (-R(n-3, x) + R(n-4, x))*M + R(n-3, x)*1_3, for n >= 3, where M^0 = 1_3 is the 3 X 3 unit matrix.
%C A321196 For the Cayley-Hamilton formula for 3 X 3 matrices with Det M = +1, sigma(3,2) = -1 and Tr(M) = x see A104578.
%C A321196 The row sums give A133872 (repeat(1, 1, 0, 0)). The alternating row sums give A057597(n+2), for n >= 0.
%C A321196 The Riordan triangle (1/(1 + x^2 + x^3), x/(1 + x^2 + x^3)) has entries t(n, m) = (-1)^(n-m)*T(n, m) (from the g.f. G(-x, -z), where the g.f. G of T is given below).
%C A321196 The inverse of Riordan T is T^{-1}, given in A321198.
%F A321196 T(n, k) = T(n-1, k-1) - T(n-2, k) + T(n-3, k), T(0, 0) = 1, T(n,k) = 0 if n < k or if k < 0. (Cf. A104578.)
%F A321196 The Riordan property T = (G(x), x*G(x)) with G(x) = 1/(1 + x^2 - x^3) implies the following.
%F A321196 G.f. of row polynomials R(n, x) is G(x, z) = 1/(1 - x*z + z^2 - z^3).
%F A321196 G.f. of column sequence k: x^k/(1 + x^2 - x^3)^(k+1), k >= 0.
%F A321196 Boas-Buck recurrence (see the Aug 10 2017 remark in A046521, also for two references):
%F A321196 T(n, k) = ((k+1)/(n-k))*Sum_{j=k..n-1} B(n-1-j)*T(j, k), for n >= 1, k = 0,1, ..., n-1, and input T(n, n) = 1, for n >= 0. Here B(n) = [x^n]*(d/dx)log(G(x)) = x*(-2 + 3*x)/(1 + x^2 - x^3) = (-1)^n*A112455(n+1), for n >= 0.
%F A321196 Recurrences from the A- and Z- sequences (see the W. Lang link under A006232 with references), which are A(n) = A321197(n) and Z(n) = A(n+1).
%F A321196   T(0, 0) = 1, T(n, k) = 0 for n < k, and
%F A321196   T(n, 0) = Sum_{j=0..n-1} Z(j)*T(n-1, j), for n >= 1, and
%F A321196   T(n, k) = Sum_{j=0..n-k} A(j)*T(n-1, k-1+j), for n >= m >= 1.
%e A321196 The triangle T(n, k) begins:
%e A321196 n\k  0   1   2   3   4   5  6  7  8  9 10 ...
%e A321196 ---------------------------------------------
%e A321196 0:   1
%e A321196 1:   0   1
%e A321196 2:  -1   0   1
%e A321196 3:   1  -2   0   1
%e A321196 4:   1   2  -3   0   1
%e A321196 5:  -2   3   3  -4   0   1
%e A321196 6:   0  -6   6   4  -5   0  1
%e A321196 7:   3  -1 -12  10   5  -6  0  1
%e A321196 8:  -2  12  -4 -20  15   6 -7  0  1
%e A321196 9:  -3  -7  30 -10 -30  21  7 -8  0  1
%e A321196 10:  5 -16 -15  60 -20 -42 28  8 -9  0  1
%e A321196 ...
%e A321196 Cayley-Hamilton formula for the matrix TS(x) =[[x,-1,1], [1,0,0], [0,1,0]] with Det(TS(x)) = +1, sigma(3, 2) = +1, and Tr(TS(x)) = x. For n = 3: TS(x)^3 = R(1, x)*TS(x)^2 + (-R(0, x) + R(-1, x))*TS(x) + R(0, x)*1_3 = x*TS(x)^2 - TS(x) + 1_3. Compare this for x = -1 with r^3 = R(3)*r^2 + (-R(2) + R(1))*r + R(2)*1 = r^2 - r + 1, where r = 1/t = A192918, with the tribonacci constant t = A058265, and R(n) = A057597(n) = R(n-2, -1).
%e A321196 Recurrence: T(5, 2) = T(4, 1) - T(3, 2) + T(2, 2) = 1 -(-1) + 1 = 3.
%e A321196 Boas-Buck type recurrence with B = {0, -2, 3, ...}:
%e A321196   T(5, 2) = ((2+1)/(5-2))*(3*1 + (-2)*0 + 0*(-3)) = 1*3 = 3.
%e A321196 Z- and A-recurrence with A(n) = {1, 0, -1, 1, -1, ...} and Z(n) = A(n+1):
%e A321196   T(4, 0) = 0*T(3, 0) - 1*T(3, 1) + 1*T(3, 2) - 1*T(3, 3) = 0 + 2 + 0 - 1 = 1.
%e A321196   T(5, 2) = 1*T(4, 1) + 0*T(4, 2) - 1*T(4, 3) + 1*T(4, 4) = 2 + 0 + 0 + 1 = 3.
%t A321196 T[n_, k_] /; 0 <= k <= n := T[n, k] = T[n - 1, k - 1] - T[n - 2, k] + T[n - 3, k]; T[0, 0] = 1; T[_, _] = 0;
%t A321196 Table[T[n, k], {n, 0, 10}, {k, 0, n}] (* _Jean-François Alcover_, Jul 06 2019 *)
%o A321196 (Sage) # uses[riordan_array from A256893]
%o A321196 riordan_array(1/(1 + x^2 - x^3), x/(1 + x^2 - x^3), 11) # _Peter Luschny_, Nov 13 2018
%Y A321196 Cf. A057597, A058265, A077961, A001005, A104578, A112455, A133872, A192918, A321197, A321198.
%K A321196 sign,easy,tabl
%O A321196 0,8
%A A321196 _Wolfdieter Lang_, Nov 09 2018